Answer
(a)
Work Step by Step
$\hat{p}=\frac{x}{n}=\frac{132}{348}\approx0.3793$
The z-value belonging to the $95\%$ confidence interval according to the table is $z=1.96$, thus the confidence interval is: $\hat{p}\pm z\sqrt{\frac{p(1-p)}{n}}$, which here is: $0.3793\pm 1.96\sqrt{\frac{0.3793\cdot(1-0.3793)}{348}}$, thus the confidence interval is $(0.328,0.43)$, thus the answer is (a)
